Cayley Graphs of Finitely Generated Groups

Cayley Graphs of Finitely Generated Groups Simon Thomas Rutgers University 13th May 2014

Cayley graphs of finitely generated groups Definition Let G be a f.g. group and let S G { 1 } be a finite generating set. Then the Cayley graph Cay(G, S ) is the graph with vertex set G and edge set E = { {x, y} y = xs for some s S S 1 }. The corresponding word metric is denoted by d S. For example, when G = Z and S = { 1 }, then the corresponding Cayley graph is: 2 1 0 1 2

But which Cayley graph? However, when G = Z and S = { 2, 3 }, then the corresponding Cayley graph is: 4 2 0 2 4 3 1 1 3 Question Does there exist an Borel choice of generators for each f.g. group such that isomorphic groups are assigned isomorphic Cayley graphs?

Standard Borel Spaces Definition A standard Borel space is a Polish space X equipped with its σ-algebra of Borel subsets. Some Examples R, [0, 1], 2 N, G fg, G,... If σ is a sentence of L ω1 ω, then is a standard Borel space. Theorem (Kuratowski) Mod(σ) = {M = N, M σ} There exists a unique uncountable standard Borel space up to isomorphism.

Borel maps Definition Let X, Y be standard Borel spaces. Then the map ϕ : X Y is Borel iff graph(ϕ) is a Borel subset of X Y. Equivalently, ϕ : X Y is Borel iff ϕ 1 (B) is a Borel set for each Borel set B Y. An Analogue of Church s Thesis EXPLICIT = BOREL

Stating the main result... Definition G fg is the space of f.g. groups with underlying set N. Theorem There does not exist a Borel map ψ : G fg N<N such that for all G, H G fg ψ(g) generates G. If G = H, then Cay(G, ψ(g)) = Cay(H, ψ(h)).

Equivalently... Definition Let C be the standard Borel space of graphs Γ with underlying set N which satisfy the following conditions: Each vertex v Γ has finite degree. Aut(Γ) acts transitively on Γ. Observation C includes the Cayley graphs of f.g. groups. Proof. If { x, xs } is an edge of the Cayley graph Cay(G, S ) and g G, then g { x, xs } = { gx, gxs }.

Equivalently... Theorem There does not exist a Borel map ϕ : G fg C such that for each G, H G fg ϕ(g) is a Cayley graph of G. If G = H, then ϕ(g) = ϕ(h). Question Why is the Theorem obviously true? Answer Because the isomorphism problem for f.g. groups is much harder than that for Cayley graphs.

Borel reductions Definition Let E, F be equivalence relations on the standard Borel spaces X, Y. E B F iff there exists a Borel map ϕ : X Y such that x E y ϕ(x) F ϕ(y). In this case, ϕ is called a Borel reduction from E to F. E B F iff both E B F and F B E. E < B F iff both E B F and E B F.

Smooth vs Nonsmooth Definition The equivalence relation E on the standard Borel space X is smooth iff there exists a Borel map ϕ : X N N such that Theorem (Folklore) x E y ϕ(x) = ϕ(y). The isomorphism relation on C is smooth. Theorem (S.T.-Velickovic) The classification problem for f.g. groups is not smooth.

Borel homomorphisms Definition The Borel map ϕ : ( X, E ) ( Y, F ) is a Borel homomorphism iff x E y = ϕ(x) F ϕ(y). Theorem If ϕ : G fg, = C, = is any Borel homomorphism, then there exist groups G, H G fg such that: ϕ(g) = ϕ(h). G and H don t have isomorphic Cayley graphs. Question But how can we be sure that two f.g. groups do not have isomorphic Cayley graphs with respect to some finite generating sets?

The basic idea of geometric group theory Although the Cayley graphs of a f.g. group G with respect to different generating sets S are usually nonisomorphic, they always have the same large scale geometry.

The quasi-isometry relation Definition (Gromov) Let G, H be f.g. groups with word metrics d S, d T respectively. Then G, H are said to be quasi-isometric, written G QI H, iff there exist constants λ 1 and C 0, and a map ϕ : G H such that for all x, y G, 1 λ d S(x, y) C d T (ϕ(x), ϕ(y)) λd S (x, y) + C; and for all z H, d T (z, ϕ[g]) C.

When C = 0 Definition (Gromov) Let G, H be f.g. groups with word metrics d S, d T respectively. Then G, H are said to be Lipschitz equivalent iff there exist a constant λ 1, and a map ϕ : G H such that for all x, y G, 1 λ d S(x, y) d T (ϕ(x), ϕ(y)) λd S (x, y); and for all z H, d T (z, ϕ[g]) = 0.

As expected Observation If S, T are finite generating sets for G, then id : G, d S G, d T is a quasi-isometry. Corollary If the f.g. groups G, H have isomorphic Cayley graphs with respect to some finite generating sets, then G, H are quasi-isometric.

A topological characterization Theorem (Gromov) If G, H are f.g. groups, then the following are equivalent. G and H are quasi-isometric. There exists a locally compact space X on which G, H have commuting proper actions via homeomorphisms such that X/G and X/H are both compact. Definition The action of the discrete group G on X is proper iff for every compact subset K X, the set {g G g(k ) K } is finite.

Obviously quasi-isometric groups Definition Two groups G 1, G 2 are said to be virtually isomorphic iff there exist subgroups N i H i G i such that: [G 1 : H 1 ], [G 2 : H 2 ] <. N 1, N 2 are finite normal subgroups of H 1, H 2 respectively. H 1 /N 1 = H2 /N 2. Proposition (Folklore) If the f.g. groups G 1, G 2 are virtually isomorphic, then G 1, G 2 are quasi-isometric.

More quasi-isometric groups Theorem (Erschler) The f.g. groups Alt(5) wr Z and C 60 wr Z are quasi-isometric but not virtually isomorphic. (In fact, they have isomorphic Cayley graphs.) Definition The f.g. group G is quasi-isometrically rigid iff whenever H is quasi-isometric to G, then H is virtually isomorphic to G. Example Z n, F n, SL n (Z),... are quasi-isometrically rigid. Question Do there exist uncountably many quasi-isometrically rigid f.g. groups?

Growth rates and quasi-isometric groups Theorem (Grigorchuk 1984 - Bowditch 1998) There are 2 ℵ 0 f.g. groups up to quasi-isometry. Proof (Grigorchuk). Consider the growth rate of the size of balls of radius n in the Cayley graphs of suitably chosen groups. Proof (Bowditch). Consider the growth rate of the length of taut loops in the Cayley graphs of suitably chosen groups.

The complexity of the quasi-isometry relation Question What are the possible complete invariants for the quasi-isometry problem for f.g. groups? Question Is the quasi-isometry problem for f.g. groups strictly harder than the isomorphism problem?

Almost there... Theorem If ϕ : G fg, = C, = is any Borel homomorphism, then there exist groups G, H G fg such that: ϕ(g) = ϕ(h). G and H are not quasi-isometric. Remark Of course, in order to prove this, we must actually show that there are many such pairs.

A little ergodic theory Definition A measure preserving action of a group Γ on a probability space (X, ν) is ergodic iff whenever Y X is a Γ-invariant Borel subset, then ν(y ) = 0, 1. Example Let µ be the usual product probability measure on 2 Γ and consider the shift action of Γ on 2 Γ = P(Γ). Then µ is Γ-invariant and Γ acts ergodically on (2 Γ, µ).

A little ergodic theory Theorem If Γ ( X, µ ) is a probability preserving action, then the following statements are equivalent. (i) Γ acts ergodically on (X, µ). (ii) If Y is a standard Borel space and f : X Y is a Γ-invariant Borel function, then there exists a Γ-invariant Borel subset M X with µ(m) = 1 such that f M is a constant function.

Some easy consequences Definition Let E Z be the orbit equivalence relation for the action of Z on 2 Z. Proposition If ψ : 2 Z, E Z N N, = is a Borel homomorphism, then there exists a Borel subset M 2 Z with µ(m) = 1 such that ψ M is a constant function. Corollary If ψ : 2 Z, E Z C, = is a Borel homomorphism, then there exists a Borel subset M 2 Z with µ(m) = 1 such that ψ maps M into a single =-class.

The heart of the matter Main Lemma There exists a Borel homomorphism such that the set θ : 2 Z, E Z G fg, = T G T { S, T 2 Z 2 Z G S, G T are not quasi-isometric } has (µ µ)-measure 1.

Proof of Theorem Suppose that ϕ : G fg, = C, = is a Borel homomorphism. Consider the composite Borel homomorphism ψ : 2 Z, E Z θ G fg, = ϕ C, =. Then there exists a Borel subset M 2 Z with µ(m) = 1 such that ψ maps M into a single =-class. Since the set { S, T 2 Z 2 Z G S, G T are not quasi-isometric } has (µ µ)-measure 1, there exist S, T M such that G S, G T are not quasi-isometric.

A slight variant of Champetier s construction Definition Let F 3 be the free group on { a, b, c } and let g Aut(F 3 ) be the automorphism defined by: g(a) = ab g(b) = ab 2 g(c) = c Lemma If w = (ac) 14, then the presentation a, b, c g n (w) n Z satisfies the C (1/6) cancellation property. Definition Let θ : 2 Z G fg be the Borel map defined by S θ G S = a, b, c R S, where R S = { g s (w) s S }.

θ is a Borel homomorphism Lemma If S E Z T, then G S = GT. Proof. Suppose that T = n + S and consider g n Aut(F 3 ). Clearly g n ( { g s (w) s S } ) = { g t (w) t T }. Hence g n induces an isomorphism from G S = a, b, c R S onto G T = a, b, c R T.

Combining ideas of Champetier and Bowditch Lemma The taut loops in the Cayley graph of G S = a, b, c R S correspond precisely to the relators R S = { g s (w) s S }. Lemma If G S, G T are quasi-isometric, then { length(g s (w)) s S } { length(g t (w)) t T } Definition If D, E N +, then D E, iff there exists k 1 such that: For every d D, there exists e E such that d/k e kd. For every e E, there exists d D such that e/k d ke.

Probability 101 Observation If l 0, then 2 l length(g l (w)) 3 l. Definition For each n 1 and k 0, let D n (k) = {l Z g k (w) /2 n g l (w) 2 n g k (w) }. Lemma There exists k 0 such that D n (k) [ k 2n, k + 2n ] [ k 2n, k + 2n ].

Probability 101 With µ-probability 1, if s 2 Z, then for each n 1, there exist infinitely many pairwise disjoint sets of the form D n (k) such that s D n (k) 0. Fix some such s 2 Z. With µ-probability 1, if t 2 Z, then for each n 1, there exist infinitely many pairwise disjoint sets of the form D n (k) such that s D n (k) 0 and t(k) = 1. It follows that G s and G t are not quasi-isometric.

The quasi-isometry relation QI is not smooth Suppose that ϕ : G fg N N witnesses that QI is smooth. Let θ : 2 Z, E Z G fg, = be a Borel homomorphism such that { S, T 2 Z 2 Z θ(s) QI θ(t ) } has (µ µ)-measure 1. Then ϕ θ : 2 Z N N is Z-invariant. Hence there exists M 2 Z with µ(m) = 1 such that (ϕ θ) M is a constant map. But then there exists S, T M M such that θ(s) QI θ(t ) and ϕ(θ(s)) = ϕ(θ(t )), which is a contradiction.

The complexity of the quasi-isometry relation Conjecture The quasi-isometry relation QI for f.g. groups is strictly more complex than the isomorphism relation = Gfg. Definition The f.g. groups G 1, G 2 are virtually isomorphic, written G 1 VI G 2, if there exist subgroups N i H i G i such that: [G 1 : H 1 ], [G 2 : H 2 ] <. N 1, N 2 are finite normal subgroups of H 1, H 2 respectively. H 1 /N 1 = H2 /N 2. Theorem (S.T.) The virtual isomorphism problem for f.g. groups is strictly harder than the isomorphism problem.

K σ equivalence relations Definition The equivalence relation E on the Polish space X is K σ if E is the union of countably many compact subsets of X X. Example The following are K σ equivalence relations on the space G fg of finitely generated groups: the isomorphism relation = the virtual isomorphism relation VI the quasi-isometry relation QI

Some universal K σ equivalence relations Theorem (Rosendal) Let E Kσ be the equivalence relation on n 1 { 1,..., n } defined by α E Kσ β N k α(k) β(k) N. Then E Kσ is a universal K σ equivalence relation. Theorem (Rosendal) The Lipschitz equivalence relation on the space of compact separable metric spaces is Borel bireducible with E Kσ.

More universal K σ equivalence relations Theorem (S.T.) The following equivalence relations are Borel bireducible with E Kσ the growth rate relation on the space of strictly increasing functions f : N N; the quasi-isometry relation on the space of connected 4-regular graphs. Definition The strictly increasing functions f, g : N N have the same growth rate, written f g, if there exists an integer t 1 such that f (n) g(tn) for all n 1, and g(n) f (tn) for all n 1.

The quasi-isometry vs. virtual isomorphism problems The Main Conjecture The quasi-isometry problem for f.g. groups is universal K σ. Theorem (S.T.) The virtual isomorphism problem for f.g. groups is not universal K σ. Corollary The virtual isomorphism problem for f.g. groups is strictly easier than the quasi-isometry relation for connected 4-regular graphs. Conjecture The virtual isomorphism problem for f.g. groups is strictly easier than the quasi-isometry relation for for f.g. groups. The End